निम्नलिखित समीकरणों को हल कीजिए -
`(i) sin theta +sin 2 theta +sin 3 theta = 1 + cos theta + cos 2 theta" "(ii)sin 4 theta - sin 2 theta = cos 3 theta`
(iii) `cos alpha - cos(alpha-theta) +cos(alpha+2 theta) = 0 " " (iv) sin^(2) n theta - sin^(2) (n-1) theta = sin^(2) theta`
(v)`sin alpha +sin (alpha+theta)+sin (alpha+2theta) = 0 " "(vi) cos 3 theta cos 5 theta cos 7 theta cos 9 theta = 0`

Answer

`(i) n pi + (-1)^(n) pi/6, 2 n pi pm (2pi)/3, " "(ii) pi/3 (2n pm 1/2) , n pi + (-1)^(n) pi/6 " "(iii) (2n pi pm pi/2) - alpha, 2n pi pm pi/3 " "(iv) r pi, pi/n (2r pm 1/2), (r pi)/(n-1)`
(v) `n pi - alpha, 2n pi pm (2pi)/3 " "(vi) (npi)/12, (npi)/4 `

Explanation

(i)कुछ सरलीकरण के पश्चात दी गयी समीकरण इस प्रकार लिख सकते है -
`(1 + cos theta + cos 2 theta)(2 sin theta-1) = 0`
यदि `2 sin theta -1 = 0 rArr theta = n pi + (-1)^(n) pi//6 " तथा जब " 1 + cos theta +cos 2 theta = 0`
`rArr theta = 2 n pi pm (2pi)/3 rArr cos theta (2 cos theta+1) = 0`